Request for crazy integrals

I'm a sucker for exotic integrals like the one evaluated in this post. I don't really know why, but I just can't get enough of the amazing closed forms that some are able to come up with.

So, what are your favorite exotic integral identities, and how do you prove them?

Here are some of my favorites: $$\int_0^\pi \sin^2\Big(x-\sqrt{\pi^2-x^2}\Big)dx=\frac{\pi}{2}$$ $$\int_0^\infty \frac{\ln(x)}{(1+x^{\sqrt 2})^\sqrt{2}}dx=0$$ $$\int_0^\infty \frac{dx}{(1+x^{1+\sqrt{2}})^{1+\sqrt{2}}}=\frac{1}{\sqrt{2}}$$ $$\int_{-\infty}^\infty \ln(2-2\cos(x^2))dx=-\sqrt{2\pi}\zeta(3/2)$$ $$\int_0^\infty \frac{\text{erf}^2(x)}{x^2}dx=\frac{4\ln(1+\sqrt{2})}{\sqrt{\pi}}$$ $$\int_0^\infty \frac{x^{3}\ln(e^x+\frac{x^3}{6}+\frac{x^2}{2}+x+1)-x^4}{\frac{x^3}{6}+\frac{x^2}{2}+x+1}=\frac{\pi^2}{2}$$ $$\int_0^{\pi/2} \ln(x^2+\ln^2(\cos(x)))dx=\pi\ln(\ln(2))$$ $$\int_0^\infty \frac{\arctan(2x)+\arctan(x/2)}{x^2+1}dx=\frac{\pi^2}{4}$$ $$\int_0^{\pi/2}\frac{\sin(x+100\tan(x))}{\sin(x)}dx=\frac{\pi}{2}$$ $$\int_0^1 \frac{x\ln(1+x+x^4+x^5)}{1+x^2}dx=\frac{\ln^2(2)}{2}$$ $$\int_0^{1/2}\sin(8x^4+x)\cos(8x^4-x)\cos(4x^2)xdx=\frac{\sin^2(1)}{16}$$

$$\int_0^{2\pi} \sqrt{2+\cos(x)+\sqrt{5+4\cos(x)}}dx=4\pi$$

And here are four extremely exotic scrumptious integrals:

$$\int_0^1 \frac{\sin(\pi x)}{x^x (1-x)^{1-x}}dx=\frac{\pi}{e}$$ $$\int_{-\infty}^\infty \frac{dx}{(e^x-x)^2+\pi^2}=\frac{1}{1+\Omega}$$

$$\int_0^\infty \frac{3\pi^2+4(z-\sinh(z))^2}{[3\pi^2+4(z-\sinh(z))^2]^2+16\pi^2(z-\sinh(z))^2}dz=\frac{1}{8+8\sqrt{1-w^2}}$$

$$\int_0^{\pi/2}\ln|\sin(mx)|\ln|\sin(nx)|dx=\frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi \ln^2(2)}{2}$$

...where $\Omega$ is the Omega Constant, $w$ is the Dottie Number, and $m,n\in\mathbb N$.

Here are some links to a few integrals: 1 (Big list, but not all of them got the right answer). From AoPS: 2, 3 , 4. Some that are solvable with Feynman's trick: here.

As for my favourites (most of them appeared on Romanian Mathematical Magazine), some are: $$I_1=\int_0^\frac{\pi}{2} \frac{\arctan(\tan x\sec x)}{\tan x +\sec x}dx=\frac{\pi}{2}\ln 2 -\frac{\pi}{6}\ln(2+\sqrt 3)$$ $$I_2=\int_0^\infty \exp\left(-\frac{3x^2+15}{2x^2+18}\right)\cos\left(\frac{2x}{x^2+9}\right)\frac{dx}{x^2+1}=\frac{\pi}{e}$$ $$I_3=\int_0^1 \frac{\ln^2 (1+x) (\ln^2 (1+x) +6\ln^2(1-x))}{x}dx=\frac{21}{4}\zeta(5)$$ $$I_4=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=-\frac{\pi}{24}$$ $$I_5=\int_0^\infty \frac{1-\cos x}{8-4x\sin x +x^2(1-\cos x)}dx=\frac{\pi}{4}$$ $$I_6=\int_0^\infty \frac{\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt{3}}-\frac{2}{3}G+\frac{\pi}{12}\ln(2+\sqrt{3})$$ $$I_7=\int_0^\infty \frac{\ln(1+x)}{x^4-x^2+1}dx=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac23 G -\frac{\pi^2}{12 \sqrt 3}$$ $$I_8=\int_0^1 \frac{\ln(1-x^2)\ln(1+x^2)}{1+x^2}dx=\frac{\pi^3}{32}-3G\ln 2+\frac{\pi}{2}\ln^22.$$ $$I_{9}=\int_0^{\frac{\pi}{4}} \ln\left(2+\sqrt{1-\tan^2 x}\right)dx = \frac{\pi}{2}\ln\left(1+\sqrt{2}\right)+\frac{7\pi}{24}\ln2-\frac{\pi}{3}\ln\left(1+\sqrt{3}\right)-\frac{G}{6}$$ $$I_{10}=\int_{-\infty}^\infty \frac{\sin \left(x-\frac{1}{x}\right) }{x+\frac{1}{x}}dx=\frac{\pi}{e^2}$$ $$I_{11}=\int_{-\infty}^\infty \frac{\cos \left(x-\frac{1}{x}\right) }{\left(x+\frac{1}{x}\right)^2}dx=\frac{\pi}{2e^2}$$ $$I_{12}=\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2} G$$ $$I_{13}=\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x}dx=\frac{\sqrt{2}\pi(5\pi^2+12\pi\ln 2 - 12\ln^22)}{96}$$ $$I_{14}=\int_0^\frac{\pi}{4} \operatorname{arcsinh} (\sin x) dx=G-\frac58\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$ $$I_{15}=\int_0^\frac{\pi}{2} x \arcsin \left(\sin x-\cos x\right)dx=\frac{\pi^3}{96}+\frac{\pi}{8}\ln^2 2$$ $$I_{16}=\int_0^\infty \int_0^\infty \frac{\ln(1+x+y)}{xy\left((1+x+y)(1+1/x+1/y)-1\right)}dxdy=\frac72 \zeta(3)$$ Where $G$ is Catalan's constant and $\operatorname{Cl}_2 (x)$ is the Clausen function.

You might find a lot of crazy integrals and series in the book, (Almost) Impossible Integrals, Sums, and Series. A few examples of integrals,

$$\int_0^{\pi/2} \cot (x) \log (\cos (x)) \log ^2(\sin (x)) \operatorname{Li}_3\left(-\tan ^2(x)\right) \textrm{d}x$$ $$ =\frac{109}{128}\zeta(7)-\frac{23}{32}\zeta(3)\zeta(4)+\frac{1}{16}\zeta(2) \zeta(5);$$ $$ \int_0^{\log(1+\sqrt{2})} \coth (x) \log (\sinh (x)) \log \left(2-\cosh ^2(x)\right)\text{Li}_2\left(\tanh ^2(x)\right) \textrm{d}x$$ $$ =\frac{73}{128}\zeta(5)-\frac{17}{64}\zeta(2)\zeta(3);$$ $$\int_0^1 \frac{\displaystyle\log^2(1-x)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x$$ $$=\frac{1}{36} \log ^6(2)-\frac{1}{6}\log ^4(2)\zeta (2)+\frac{7}{24} \log ^3(2) \zeta (3)+\frac{5}{8}\log ^2(2) \zeta (4)-\frac{581}{48} \zeta (6)$$ $$ -\frac{7}{8} \log (2) \zeta (2)\zeta (3)-\frac{79}{64} \zeta^2 (3);$$

$$ \sin (\theta)\sin\left(\frac{\theta}{2}\right)\int_0^1 \frac{\displaystyle x}{(1-x) \left(1-2 x \cos (\theta)+x^2\right)} (\zeta (m+1)-\text{Li}_{m+1}(x)) \textrm{d}x$$ $$ =(-1)^{m-1} \sum_{k=1}^{\infty}\frac{H_{k+1}}{(k+1)^{m+1}}\sin\left(\frac{k \theta}{2}\right)\sin\left(\frac{(k+1)\theta}{2}\right)$$ $$ +(-1)^{m-1}\sum_{i=2}^{m} (-1)^{i-1}\zeta(i)\sum_{k=1}^{\infty}\frac{\displaystyle \sin\left(\frac{k\theta}{2}\right)\sin\left(\frac{(k+1) \theta}{2}\right)}{(k+1)^{m-i+2}};$$ $$\sin\left(\frac{\theta}{2}\right)\int_0^1\frac{x(\cos(\theta)-x)}{(1-x)(1-2x\cos(\theta)+x^2)}(\zeta (m+1)-\text{Li}_{m+1}(x))\textrm{d}x$$ $$ =(-1)^{m-1}\sum_{k=1}^{\infty}\frac{H_{k+1}}{(k+1)^{m+1}}\sin\left(\frac{k\theta}{2}\right)\cos\left(\frac{(k+1)\theta}{2}\right)$$ $$ +(-1)^{m-1}\sum_{i=2}^{m}(-1)^{i-1} \zeta(i)\sum_{k=1}^{\infty} \frac{\displaystyle \sin\left(\frac{k\theta}{2}\right)\cos\left(\frac{(k+1)\theta}{2}\right)}{ (k+1)^{m-i+2}}.$$

A few examples of series (which you may also transform into some fancy integrals if you wish to),

$$\sum_{n=1}^{\infty}\frac{H_n}{n^2}\left(\frac{ H_1}{1^3}+\frac{H_2}{2^3}+\cdots +\frac{H_n}{n^3} \right)=10\zeta(7)+\frac{9}{2}\zeta(2)\zeta(5)-\frac{23}{2}\zeta(3)\zeta(4);$$ $$ \sum_{n=1}^{\infty}\frac{H_n}{n^3}\left(\frac{H_1}{1^2}+\frac{H_2}{2^2}+\cdots +\frac{H_n}{n^2} \right)=\frac{23}{2}\zeta(3)\zeta(4)-\frac{11}{2}\zeta(2)\zeta(5)-4\zeta(7);$$ $$\sum_{n=1}^{\infty}\frac{H_n^2}{n^2}\left(\frac{H_1}{1^2}+\frac{H_2}{2^2}+\cdots +\frac{H_n}{n^2} \right)=\frac{45}{16}\zeta(7)-\frac{7}{2}\zeta(2)\zeta(5)+\frac{17}{2}\zeta(3)\zeta(4);$$ $$\sum_{n=1}^{\infty}\frac{H_n}{n^2}\left(\frac{H_1^2}{1^2}+\frac{H_2^2}{2^2}+\cdots +\frac{H_n^2}{n^2} \right)=\frac{93}{8} \zeta(7)+\frac{11}{2}\zeta(2)\zeta(5)-\frac{51}{4}\zeta(3)\zeta(4);$$ $$ \zeta(4)$$ $$ =\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{ \left(H_{2 n}\right)^2}{ (2 n+1)^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}}{(2 n+1)^3}$$ $$ -\frac{8}{5}\sum _{n=1}^{\infty } \frac{\left(H_{2 n}\right){}^2}{ n^2}-\frac{32}{5}\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n+1)^2}-\frac{64}{5}\log(2)\sum _{n=1}^{\infty } \frac{H_{2 n}}{(2 n+1)^2}-\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{ n^2}.$$

Extremely crazy integrals you may also find in the paper The derivation of eighteen special challenging logarithmic integrals by Cornel Ioan Valean.

I'm sure a lot of crazy integrals you'll also meet in the sequel of the book (Almost) Impossible Integrals, Sums, and Series since the author prepares a continuation of this book.

$$\int_{-\infty}^\infty\prod_{k=1}^n\operatorname{sinc}{\theta\over(2k-1)}d\theta=\pi ,$$provided $n\in{1 ... 7}$ ... for $n\geq8$, it starts being $<π$ by the most miniscule amounts!

I'm partial to the one in this question What is the Centroid of $z=\frac{1}{(1-i\tau)^{i+1}},\ \ \tau\in (-\infty,\infty)$ .

I found a solution, but it was hardly elegant. A solution that doesn't use hypergeometric functions in the middle of the solution would be nice.